Let $\mathbf A=\left[\begin{matrix}\mathbf A_{11}&\mathbf A_{12}\\ \mathbf A_{21}&\mathbf A_{22}\end{matrix}\right]$ be a positive semi-definite matrix, $\mathbf A_{ij}\in\mathbb C^{n\times n}$ and $rank(\mathbf A)=n$.
Prove or disprove that
$$
\lambda_{max}(\mathbf A) + \lambda_{max}(\mathbf A_{11}+\mathbf A_{22}) \leq
\lambda_{max}\left(\left[\begin{matrix}tr(\mathbf A_{11})&tr(\mathbf A_{12})\\ tr(\mathbf A_{21})&tr(\mathbf A_{22})\end{matrix}\right]\right) + tr(\mathbf A_{11}+\mathbf A_{22}),
$$
where $\lambda_{max}$ denotes the largest eigenvalue and $tr$ denotes the trace.
I tried small values of $n$. For $n=1$ we have "$=$" (the same terms in the LHS and RHS). I have a rather long proof for $n=2$ (in my proof I also replaced $\mathbb C$ by $\mathbb R$).
It seems that some results on (completely) positive maps
are somehow related to the inequality ( indeed, all the maps
$$
\mathbf A\rightarrow \mathbf A_{11}+\mathbf A_{22},\qquad
\mathbf A\rightarrow \left[\begin{matrix}tr(\mathbf A_{11})&tr(\mathbf A_{12})\\ tr(\mathbf A_{21})&tr(\mathbf A_{22})\end{matrix}\right],\qquad
\mathbf A\rightarrow
tr(\mathbf A_{11}+\mathbf A_{22})
$$
are positive). By Theorem 2.3.7 (The Russo-Dye Theorem) from the book R. Bhatia, Positive definite matrices,
$$
\lambda_{max}\left(\left[\begin{matrix}tr(\mathbf A_{11})&tr(\mathbf A_{12})\\ tr(\mathbf A_{21})&tr(\mathbf A_{22})\end{matrix}\right]\right)\leq n\lambda_{max}(\mathbf A),
$$
which is useless since we need inequality of the type $\lambda_{max}(\mathbf A)\leq\dots$
Best Answer
Write $A=\sum w_i\otimes w_i$ where $w_i$ are orthogonal and $\|w_1\|=\max_i\|w_i\|$. Also write $w_i=(u_i,v_i)$. Then the inequality in question is just $$ \|u_1\|^2+\|v_1\|^2+\max_{e:\|e\|=1}\sum_i[|(u_i,e)|^2+|(v_i,e)|^2] \\ \le \left\|\begin{pmatrix}\sum_i\|u_i\|^2&\sum_i(u_i,v_i)\\\sum_i(v_i,u_i)&\sum_i\|v_i\|^2\end{pmatrix}\right\|+\sum_i(\|u_i\|^2+\|v_i\|^2) $$ Now, cancel the first term with the corresponding term on the RHS and estimate $ \sum_{i\ge 2}[|(u_i,e)|^2+|(v_i,e)|^2] $ by $ \sum_{i\ge 2}(\|u_i\|^2+\|v_i\|^2) $. Also, note that the norm of the sum of several positive definite matrices is not less than the norm of each of them, so we can remove all $i\ge 2$ in the matrix and reduce the problem to showing that $$ |(u_1,e)|^2+|(v_1,e)|^2\le \left\|\begin{pmatrix}\|u_1\|^2&(u_1,v_1)\\(v_1,u_1)&\|v_1\|^2\end{pmatrix}\right\| $$ However, we can write the matrix on the right as the sum of positive definite matrices of the kind $\begin{pmatrix}|(u_1,e_j)|^2& (u_1,e_j)(e_j,v_1)\\(v_1,e_j)(e_j,u_1)&|(v_1,e_j)|^2\end{pmatrix}$ where $e_j$ is an orthonormal basis with $e_1=e$. Again, we can remove all matrices with $j\ge 2$, after which we get an identity.